This is a NE555-based timer circuit to simultaneously control two independant relays, with a momentary push-button starting a roughly 10 minute "on-time."
D1 is sort of redundant, but I want to keep it in the circuit strictly as a protective measure for the NPN transistor Q1.

I absolutely welcome any critiques you may have, but please be gentle, this is my first time!

I'm an audio engineer (commercial system design), not an electronics engineer, and oddly enough, THIS design is for my race car...

Thanks!

Dave

 

I think the symbols of the relays are wrong. I don't understand where are the coils.

 

It appears the coil is labeled 86 and 0. I'm assuming the 12V is switched.

I haven't redrawn it yet.

 

I think the symbols of the relays are wrong. I don't understand where are the coils.

You may well be correct... the terminal labeling is direct from a Bosch 12V 30A relay, with a coil impedance of 88 ohms... Terminal 85 is the 12V supply, 86 is the coil lead (N/O), 87 is the switched 12V lead to the controlled device, and terminal 0 is ground for the coil. How do you draw a relay in a circuit?

 

Dave,

I'm curious as to why your output is arranged in such a manner.

In your title block you state the circuit is "on" for approximately 10 minutes. That isn't so, it's more like 320 seconds or 5.2 minutes.

 

It appears the coil is labeled 86 and 0. I'm assuming the 12V is switched.

I haven't redrawn it yet.

Absolutely! The relays switch on a 12V load. Thanks for taking the time to look at this...

 

Dave,

I'm curious as to why your output is arranged in such a manner.

In your title block you state the circuit is "on" for approximately 10 minutes. That isn't so, it's more like 320 seconds or 5.2 minutes.

I haven't breadboarded this yet, but isn't there quite a bit of leakage with electrolytic caps? I was anticipating that to extend the amount of time that the 555 output is high. If needs be, I'll stick a variable in place of R3, play until I get around 10 minutes or so, and then measure the value.

What about the output is odd? I went with a transistor to act as a current amplifier to drive the relay coils. If the circuit only used one relay, then the 136mA draw could be directly supplied by the NE555 (200mA), but the two coils would need 272mA, which is beyond the capability of the chip to supply. D1 is in place to protect the transistor from any voltage spike as the relays open, and is redundant protection for the primary diodes, D2 and D3 at the relays themselves.

I'm wide open to suggestions... I'm a pure novice at this, drawing on a couple of classes from 15 years ago, some online research, and an idea.

 

You may well be correct... the terminal labeling is direct from a Bosch 12V 30A relay, with a coil impedance of 88 ohms... Terminal 85 is the 12V supply, 86 is the coil lead (N/O), 87 is the switched 12V lead to the controlled device, and terminal 0 is ground for the coil. How do you draw a relay in a circuit?

That explains it. Normally corporations don't worry with the kind of details that we do.

I haven't breadboarded this yet, but isn't there quite a bit of leakage with electrolytic caps? I was anticipating that to extend the amount of time that the 555 output is high. If needs be, I'll stick a variable in place of R3, play until I get around 10 minutes or so, and then measure the value.

Tantalum types are better for this kind of application. They have a much smaller leakage current than regular alluminium electrolytic caps. As a matter of fact, it is quite normal to find tantalum caps in projects using 555's (as time base capacitors).

What about the output is odd? I went with a transistor to act as a current amplifier to drive the relay coils. If the circuit only used one relay, then the 136mA draw could be directly supplied by the NE555 (200mA), but the two coils would need 272mA, which is beyond the capability of the chip to supply. D1 is in place to protect the transistor from any voltage spike as the relays open, and is redundant protection for the primary diodes, D2 and D3 at the relays themselves.

If I'm seeing it right, D1 won't protect the transistors from the back emf generated by the coils. As you may see, there is no alternate low resistance path from pin 86 of RL2 to pin 0 of RL1. Same applies to RL2. Thus, the transistor will suffer the consequences.

However, you can use only one diode to protect the transistor from the back emf generated by the coils. Assuming that you are connecting both coils in paralel, you can use one diode backwards from pin 86 to pin 0.
Also I noticed that the coils should be connected to Vcc and not to ground. D1 should be taken off or will short the transistor. Also, D2 and D3 serve no purpose.

A final note: you should put "Copyleft...All rights reversed" instead of "Copyright...All rights reserved"

 

Dave,

After you correct what cumesoftware stated, post your revised drawing.

The on Time is calculated by 1.1 * R * C

If you pick your C, then R would be Time / 1.1 * C
If you picked your R, then C would be Time / 1.1 * R

I'll be waiting for your revised drawing, as I have already revised mine.

A couple of questions ... what is the relay's OEM number?

How much current are you delivering to the load ... I know the contacts are rated at 20 amperes.

 

How do you draw a relay in a circuit?

Here is a link that has one version. There are other variations but they usually all show a coil and the switches.

http://www.mitedu.freeserve.co.uk/Prac/symbol.htm

 

Dave,

After you correct what cumesoftware stated, post your revised drawing.

The on Time is calculated by 1.1 * R * C

If you pick your C, then R would be Time / 1.1 * C
If you picked your R, then C would be Time / 1.1 * R

I'll be waiting for your revised drawing, as I have already revised mine.

A couple of questions ... what is the relay's OEM number?

How much current are you delivering to the load ... I know the contacts are rated at 20 amperes.

First off, thank to BOTH of you for your help! After looking at your comments, scratching my head, and then looking again, I can see that I completely duffed the transistor-relay portion of the design! Attached is a revision... I am, however, going to stick with using a standard electrolytic cap in the final design, primarily due to cost. In the end the "10 minute" is an approximation, and is not materially critical to the success of the finished piece. A 20% 470uF cap has a cost of around .12 USD, but a 10% tantalum piece has a cost of about $42USD!!

The "Bosch relay" in the automotive world is a genericised description of a four-terminal SPST NO relay, and as such has no specific OE part number. An example can be found here.

In the meantime, please find attached my latest revision, showing the corrected cap and resistor values for a "true" 10 minute control circuit. In the final piece, I will probably wind up popping in a pot, and then measuring the actual resistance needed to give me the ballpark 10 minutes.

The actual application is for control of two items in my car, in between races. The first is an electric water pump, and the second is the coolant fan. My idea is to be able to pull some heat out of the engine for more consistency, but I tend to be easily distracted at the track, and would hate to come back to a dead battery, having forgotten to turn off the cool-down circuit...

The water pump has an approximate draw of 12.5A, and the fan about 11A The relay in my design will simply add an alternative voltage source in between the ignition-switched relays and their loads, allowing power to be applied without the ignition on. Hopefully that made sense...

I look forward to comparing the redesign with yours!

 

Thank you so much for your help with this! I do sincerely appreciate it!

 

Just one little nitpick... I'd recommend a small cap on Pin 5 of the NE555. A 0.1uF would be sufficient.

Where will this circuit be located on the car? Will it be exposed to extreme temperature? Al Electrolytics have a temperature coefficient and lose capacitance at lower temps -- 20%. Higher temps reduce their life-- 10deg increase halves it. These are not major issues for your application but things to consider. You can put a tantalum in parallel which would help keep the capacitance stable over temp. If you do then please derate the voltage of the tantalum by 50% (25V part). Look for a 105C rated Al electrolytic cap (16V is OK)

edit:
On start-up you might see a long glitch on the output since the threshold has to charge up.

 

First off, thank to BOTH of you for your help! After looking at your comments, scratching my head, and then looking again, I can see that I completely duffed the transistor-relay portion of the design! Attached is a revision... I am, however, going to stick with using a standard electrolytic cap in the final design, primarily due to cost. In the end the "10 minute" is an approximation, and is not materially critical to the success of the finished piece. A 20% 470uF cap has a cost of around .12 USD, but a 10% tantalum piece has a cost of about $42USD!!!

Well, cheched your schematic and it is almost OK. One of the diodes, D1 or D2 is redundand. You will only need one diode there.
It seems a fact that tantalum caps are not cheap. I only used small capacity ones. But in your case, since your capacitor is relatively big, you can substitute it by a regular aluminum electrolytic one. Leakage is only a relevant problem with small caps of up to 10uF.

One last advice. Breadboard the circuit first to see if everything is working. I always do a test and found many faults that way (namely in audio amplifiers).

 

Just one little nitpick... I'd recommend a small cap on Pin 5 of the NE555. A 0.1uF would be sufficient.

Where will this circuit be located on the car? Will it be exposed to extreme temperature? Al Electrolytics have a temperature coefficient and lose capacitance at lower temps -- 20%. Higher temps reduce their life-- 10deg increase halves it. These are not major issues for your application but things to consider. You can put a tantalum in parallel which would help keep the capacitance stable over temp. If you do then please derate the voltage of the tantalum by 50% (25V part). Look for a 105C rated Al electrolytic cap (16V is OK)

My intention is to mount the control under the dash. I would anticipate very little use in colder weather (<45*F), but the upper temps could be a factor, it can get HOT in the car (~130*F on a long run). Thanks for the tip on the high-temp caps...

edit:
On start-up you might see a long glitch on the output since the threshold has to charge up.

I must be missing something... I thought that as long as the voltage applied to pin 6 (threshold) was below 2/3 of the circuit voltage (around 8V in this case), the output state of the chip would be high, i.e. 12V. When the cap charged up enough to bring the voltage on that leg above 2/3, then the output of the chip (pin 3) would go low, or 0V in this case. Is this not accurate? If it's not, where am I going wrong? I really appreciate your thoughts and guidance in this.

edit: Why are you suggesting adding a small-value cap to the control pin? Just to provide some signal isolation?

 

Well, cheched your schematic and it is almost OK. One of the diodes, D1 or D2 is redundand. You will only need one diode there.
It seems a fact that tantalum caps are not cheap. I only used small capacity ones. But in your case, since your capacitor is relatively big, you can substitute it by a regular aluminum electrolytic one. Leakage is only a relevant problem with small caps of up to 10uF.

One last advice. Breadboard the circuit first to see if everything is working. I always do a test and found many faults that way (namely in audio amplifiers).

First off, thank you for taking the time to help with this...

As for the diode redundancy, do you see any actual design issues with this? I'm planning on mounting the relays in the engine compartment, and the timer circuit under the dash, not all in one chassis box, and would like the peace of mind knowing that there would be no glitches. I'm thinking in terms of a wiring fault somewhere that would wind up opening the feed line to one of the relays and removing the protection from the circuit.

 

I must be missing something... I thought that as long as the voltage applied to pin 6 (threshold) was below 2/3 of the circuit voltage (around 8V in this case), the output state of the chip would be high, i.e. 12V. When the cap charged up enough to bring the voltage on that leg above 2/3, then the output of the chip (pin 3) would go low, or 0V in this case.

Yes you are correct. I thought you did not want the output to be high until you pressed the button.

The small cap is for noise filtering. The control pin is hooked directly up one of the internal comparators and transients could accidently trigger it.

Do you have bypass caps on the power input Pin 8 to ground. A 0.1uF ceramic and a 2.2uF tantalum (or higher) in parallel.

I'd also put some TVS diodes on the 12V and maybe across the BJT collector to ground.